6log(x^2+1)-x=0
I will assume that your base is 10
log(x^2 + 1) = x/6
x^2 + 1 = 10^(x/6)
There is no easy way to do this, so I will resort to good ol' Wofram, (one of the best math-sites on the net)
http://www.wolframalpha.com/input/?i=solve+x%5E2+%2B+1+%3D+10%5E%28x%2F6%29
as you can see
x = 0
x = .416103
x = 3.6267
I tested all three answers , they all work
I had done this question before for somebody else.
However, in that solution I assumed the base was e
and of course we got 3 different answers
http://www.jiskha.com/display.cgi?id=1365053187
I'm sorry, that equation is way too serious for a goofy clown like me. I specialize in laughter, not logarithms. Maybe you can ask me a joke instead?
To solve the equation 6log(x^2+1)-x=0, follow these steps:
Step 1: Move the x term to the right side of the equation:
6log(x^2+1)=x
Step 2: Divide both sides of the equation by 6:
log(x^2+1)=x/6
Step 3: Rewrite the logarithmic equation in exponential form:
x^2+1=10^(x/6)
Step 4: Subtract 1 from both sides of the equation:
x^2=10^(x/6) - 1
At this point, it is not possible to find an exact solution for x. However, you can use numerical methods (such as graphing or iterative methods) to estimate the solution.
To solve the equation 6log(x^2 + 1) - x = 0, follow these steps:
Step 1: Move the x term to the right side of the equation.
6log(x^2 + 1) = x
Step 2: Eliminate the logarithm by applying the exponential function.
Rewrite the equation using the exponential form of logarithm:
x^2 + 1 = 10^(x/6)
Step 3: Rearrange the equation and express it as a polynomial equation.
x^2 - 10^(x/6) + 1 = 0
Step 4: Solve the equation using numerical or graphical methods.
Unfortunately, there isn't a simple algebraic way to solve this equation. You can use numerical methods, such as Newton's method or graphical methods, to approximate the solution.
If you are using a graphing calculator or a math software, you can also graph the function y = 6log(x^2 + 1) on one side of the equation and y = x on the other side. Then, find the point(s) of intersection of the two graphs, which would give you the approximate solutions to the equation.
Remember to check if the solutions satisfy the original equation as well.